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.9999........ is not a number
have you taken calculus?
scroll down to number 4, its actaully an easy proof.
you will enjoy calculus:)
Here is an algebraic proof of it: 10x = 9.999999... with x = 0.99999... 10x - x = 9.99999... - 0.999999... 9x = 9 Thus, x = 1
This question was asked because I was trying to exaplin that inbetween any two numbers is another number, he then said what about .9999999.... and 1, i then tried to explain to him that .99999999.. was not a finite number, then we had this question posted:)
there is no amount of 9's that you can write after .9 that will make it equal to 1, but the limit as the amount of 9's goes to infinity is said to equal 1
wait... doesn't 0.9999... = 9/10 + 9/100 + 9/1000 + .... an infinite bounded series?
The number "0.9999..." can be "expanded" as: 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...
So the formula proves that 0.9999... = 1.
but .99999...is not a number is the point of this topic
0.999999..... is a no. we have tp prove that it is equal to 1
its a number if you think about convergence, but you can never write 1 in the form .999999999999, read the last question and you will see the point im trying to make
its not a finite number.... it can be looked at as a sequence....at infinity then its a number. I told him that between any two numbers is another number, he then said what about .99999999999..... and 1, this is what followed.
Its an extended real number in the since that infinity is an extended real number....
"But", some say, "there will always be a difference between 0.9999... and 1." Well, sort of. Yes, at any given stop, at any given stage of the expansion, for any given finite number of 9s, there will be a difference between 0.999...9 and 1. That is, if you do the subtraction, 1 – 0.999...9 will not equal zero. But the point of the "dot, dot, dot" is that there is no end; 0.9999... is inifinte. There is no "last" digit. So the "there's always a difference" argument betrays a lack of understanding of the infinite. (That's not a "criticism", per se; infinity is a messy topic.)
right and i dont think he has had calculus...
this started with density of rational/irational and archimedes principle on the real line
the point is, that there is an infinite amount of numbers between any two numbers a,b where a != b
hahhaha.. we are just fighting you and me are absolutely right...it's a skull problem..right??
This one might make him want to quit math The infinite amount of numbers between 0,1 is larger than the infinite amount of numbers between 0 and infinity.:)
0.9999 is approximately equal to 1 but it is not exactly equal.. \[0.999 \approx 1 \qquad \qquad (0.999 \ne 1)\]
can you provide a proof of this ... "This one might make him want to quit math The infinite amount of numbers between 0,1 is larger than the infinite amount of numbers between 0 and infinity.:)"
no, my analysis teacher said it two days ago. she said the elemnts between 0 and 1 have more mappings than 0-infinity
your teacher is either wrong or you have misquoted her
the real numbers between 0 and 1 is the same size as the real numbers from 0 to infinity
\[1 - 0.999\cdots = 0.000\cdots = 0\]